A ball is dropped 2.75 m to the floor below. What is the acceleration of the ball while it is on the floor if it is on the floor for 0.0400 s?
I'm really confused on this question because it is a free fall question so wouldn't the acceleration be -9.81 m/s/s? How is the answer +340.0 m/s/s?
This is what I have but I don't know which equation to use...
y = -2.75 m
vi = 0 m/s
vf = ? - m/s
a = -9.81 m/s/s
t = 0.0400 s
You have the right idea about getting the speed at which it hits the floor.
HOWEVER the velocity changes when it hits
acceleration = change of velocity / change in time
it is on the floor, so velocity goes from contact speed to zero in 0.04 seconds
now
contact speed:
a = -9.81 m/s^2
v = 0 - 9.81 t
0 = 2.75 + 0 - 4.9 t^2
so t = sqrt (2.75 / 4.9) = 0.749 seconds
so v = - 9.81* 0.749 = -7.35 m/sec t contact
NOW do the problem
a = [0 - ( -7.35) ] / time on floor (positive up)
Now of course I assumed the ball stopped. If it is a perfect ball and bounced UP at 7.35 m/s , then double the result because the change in velocity doubled
I tried it but I keep on getting the answer 367.5 m/s/s but the answer is +340.0 m/s/s what am I doing incorrectly?
Well, this is quite a puzzling question, isn't it? It's understandable that you're confused.
In this case, since the ball is on the floor for only a very short time, we can consider it as a collision. During the collision, there is a change in momentum, and that change happens very quickly, resulting in a large acceleration.
So, the acceleration of +340.0 m/s/s is actually the average acceleration during the extremely short time interval when the ball is in contact with the floor. It's important to note that the acceleration can indeed change during the collision.
Keep in mind, though, that it's just a temporary and rather extreme acceleration caused by the collision. In the overall motion of the ball's free fall, the acceleration is still -9.81 m/s/s.
Physics can be quite clownish sometimes, but I assure you, it all adds up in the end!
To find the acceleration of the ball while it is on the floor, we can use the kinematic equation:
y = vi * t + (1/2) * a * t^2
In this equation, y represents the displacement (in this case, the height) of the ball, vi represents the initial velocity (which is 0 m/s since the ball is dropped from rest), a represents the acceleration, t represents the time, and t^2 represents the squared time.
In this scenario, the ball is dropped 2.75 m, so y = -2.75 m (negative because it is downward). We are given that t = 0.0400 s, and we want to find the acceleration, which is a.
Plugging in the values, we get:
-2.75 m = 0 m/s * 0.0400 s + (1/2) * a * (0.0400 s)^2
Simplifying the equation, we find:
-2.75 m = (1/2) * a * 0.0016 s^2
Now, we can solve for the acceleration by isolating the variable a:
a = (-2.75 m) / ((1/2) * 0.0016 s^2)
a = (-2.75 m) / 0.0008 s^2
This gives us:
a = -3437.5 m/s^2
So, the acceleration while the ball is on the floor is indeed -3437.5 m/s^2. The positive answer of +340.0 m/s^2 you mentioned seems to be a typo or a mistake in the given answer. The negative sign indicates that the acceleration is directed downward, which is the natural direction of free fall due to gravity.